Question

If ${V_o}$ be the orbital velocity of a satellite in a circular orbit close to the earth’s surface
and ${V_e}$ is the escape velocity from the earth, then relation between the two is
A. ${V_e} = 2{V_o}$
B. ${V_e} = \sqrt 3 {V_o}$
C. ${V_e} = {V_o}\sqrt 2 $
D. ${V_o} = {V_e}$

Answer 1

Hint: Orbital velocity is the velocity with which an object revolves around a huge body and escape velocity is the velocity required by an object to overcome the gravitational force of the huge body and escape from the orbit to infinity. Write the formulas of orbital velocity and escape velocity and compare them to find a relation between them.

Complete step by step answer:
Orbital velocity of a satellite is ${V_o}$ and escape velocity of the satellite is ${V_e}$
We are given to find the relation between escape velocity and the orbital velocity of the satellite from the earth.
For a spherically symmetric huge body such as a satellite, the escape velocity, at a given distance is
${V_e} = \sqrt {\dfrac{{2GM}}{R}} $
Orbital velocity of a satellite in a circular orbit close to the earth’s surface is
${V_o} = \sqrt {\dfrac{{GM}}{R}} $
Escape velocity of the satellite can be written as ${V_e} = \sqrt {\dfrac{{2GM}}{R}} = \sqrt 2 \times \sqrt {\dfrac{{GM}}{R}} $
Where $\sqrt {\dfrac{{GM}}{R}} $ is the orbital velocity ${V_o}$. Substitute ${V_o}$ in the place of $\sqrt {\dfrac{{GM}}{R}} $ in the above equation.
The resulting equation will be
$
{V_e} = \sqrt 2 \times {V_o} \\
{V_e} = {V_o}\sqrt 2 \\
$
The escape velocity of the satellite is square root of 2 times of its orbital velocity.
Therefore, the correct one is Option C, ${V_e} = {V_o}\sqrt 2 $

Note: Orbital velocity is the velocity needed to keep a satellite moving in an orbit whereas critical velocity is the velocity required to put the same satellite into an orbit. There is a slight difference between these two velocities. Do not confuse while defining them.